Why can different integration variables appear on different sides of an equation?

Question

I am a student of physics and I am trying to wrap my head around the following:

$$\omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}$$

implies

$$\Delta \theta = \int \omega\,\mathrm{d}t$$

So I understand from the first relation we have $\mathrm{d}\theta = \omega\,\mathrm{d}t$, but how do you integrate both sides to get a $\Delta \theta$ on the left side of the second relation? Wouldn't you have to integrate from $\theta_i$ to $\theta_f$? But it seems to me you would then have to use those same limits on the right side integral — but the integral should be integrating over time, not theta.

The reason I am wondering about this is I am solving a problem about average angular velocity and it involves finding the area under the curve of a $\omega(t)$ function. I know I need to use that second relation but I don’t see how it follows from the first. Thank you!

Answer 1

You are saying that $\Delta\theta = \theta_{\rm j} - \theta_{\rm i}$

Well lets start with $\displaystyle \int_{t_{\rm i}}^{t_{\rm f}} \omega \,dt$ and substitute $\omega = \dfrac {d\theta}{dt}$ into that integral

which gives $\displaystyle \int_{t_{\rm i}}^{t_{\rm f}} \left (\dfrac {d\theta}{dt} \right) \,dt$ and this is the same as $\displaystyle \int_{\theta_{\rm i}}^{\theta_{\rm f}} d\theta = \theta_{\rm f} - \theta_{\rm i} = \Delta \theta$

where $\theta = \theta_{\rm i}$ when $t = t_{\rm i}$ and $\theta = \theta_{\rm f}$ when $t = t_{\rm f}$ the limits of integration having to change because $t$ was the old variable and now $\theta$ is the new variable.

Answer 2

Formally you have $\Delta\theta=\int_{\theta_i}^{\theta_f}d\theta=\int_{t_i}^{t_f}\omega dt$, where the last equation is obtained by integration by substitution. Physicists often don't bother showing integration limits on a definite integral if it's obvious what they will be. In this case, they also haven't written $\Delta\theta$ as an integral.

Answer 3

Lets take the simpler example of $v = \frac{{\rm d}x}{{\rm d}t}$ which is solved for $ x = \int {\rm d}x = \int v {\rm d} t$ and look at why the two integrals by different variables must be equal.

• Lemma 1 - The calculus here involves only one independent variable, and all other variables are depended on this one variable.

So an infinitesimal change in position is defined as

$$ {\rm d}x = \frac{\partial x}{\partial t}\, {\rm d}t = v \,{\rm d}t$$

with the assumption that for small time frames $v$ is constant. In this context, the two integrals below are equivalent, because the individual little changes are equal to each other, as long as the limits correspond also.

$$ x = \int {\rm d}x = \int v \,{\rm d}t$$

But you can only do this if you have a single independent quantity.

Consider a system where $v(x,t) = V \exp(-\beta t) \sin(k x)$. Then

$${\rm d}v = \frac{\partial v}{\partial t}{\rm d}t + \frac{\partial v}{\partial x}{\rm d}x = V \exp(-\beta t) \left( k \cos(k x){\rm d}x - \beta \sin(k x) {\rm d}t \right) $$

or

$$ v = \int {\rm d}v = V \left( \int k \exp(-\beta t) \cos(k x)\,{\rm d}x - \int \beta \exp(-\beta t) \sin(k x)\,{\rm d}t \right) $$

which cannot be solved directly unless we can relate $x$ with $t$ making it a one independent variable problem.